1. Prof. Yuan-Shyi Peter Chiu
Materials Management
2017/4/26
Material Management
Class Note # 5
SCHEDULING
Prof. Yuan-Shyi Peter Chiu
Feb
Dr. Yuan-Shyi Peter Chiu

2. ◇ § S1：Introduction Materials Management 2017/4/26
Dr. Yuan-Shyi Peter Chiu

3. § S1：Introduction ◇

4. ◇ § S5：Characteristics of Job Shop Scheduling Materials Management
2017/4/26 ◇
§ S5：Characteristics of Job Shop Scheduling
Dr. Yuan-Shyi Peter Chiu

5. ◇ § S5：Objectives of Job Shop Management
IT IS IMPOSSIBLE TO OPTIMIZE ALL 7 OBJECTIVES SIMULTANEOUSLY.
 CUSTOMER Service Quality vs.  Plant Efficiency / Cost

6. ◇ § S6：Flow Shop vs. Job Shop ... ...   (1) FLOW SHOP
job M/C M/C M/C M/C M/C 1 2 3 4 m     . 
... n
Assembly Line
different M/C = different operations.
(2) JOB SHOP
job M/C M/C M/C M/C M/C
... 1 2 3 4 m   .  n
{1, 3, 4, 1, m}
{2, 4, 2, m-1}
PROBLEMS are extremely complex.
All-purpose solution algorithms for solving general job shop problems do not exist.

7. ◇ § S6：Flow Shop vs. Job Shop ... ...  
(3) Parallel processing vs. Sequential processing
JOB M/C M/C M/C M/C
... 1 2 3 m
Sequential
  .  n
JOB M/C M/C M/C M/C    1 3 2 m 1 3
Parallel processing
... 2 m 1 3  .  n 3

8. ◇ § S7：Indicators of Performance Evaluation job 1 t1 = 25 sec.
(1) FLOW TIME
job t1 = 25 sec.
job t2 = 13 sec.
job t3 = 21 sec. .
(2) MEAN FLOW TIME

9. ◇ § S7：Indicators of Performance Evaluation
(3) MAKESPAN：F[n]
TIME REQUIRED TO COMPLETE ALL n jobs
Minimizing to the Makespan is a common objective
in multiple - m/c sequencing problem.
(4) TARDINESS
max ( F[i] - D[i] , 0 ) where F[i] : Completion time of job i.
D[i] : Job i ’s Due Date.
Example ：
Job F[i] D[i] Tardiness

10. ◇ § S7：Indicators of Performance Evaluation (5) LATENESS F[i] - D[i]
Job F[i] D[i] Lateness
(6) MINIMIZING AVG. TARDINESS
MAX. TARDINESS
ARE COMMON SCHEDULING OBJECTIVES.

11. ◇ § S8：Notation ti ; di  Wi = Fi － ti  Fi = Wi + ti Ti = max[Li , 0]
  Li = Fi － di
Ti = max[Li , 0]
Ei = max[-Li , 0]
Tmax = max {T1 , T2 , … , Tn}
→ SPT minimizes
Mean flow Time.
Mean waiting Time.
Mean lateness.
→ EDD minimizes maximum lateness Lmax ～ Tmax.
Lmax = max {L1 , L2 , L3 , … , Ln}

12. ◇ Common Scheduling Rules for single machine: FCFS
Shortest Processing Time (SPT)
(3) Earliest Due Date (EDD)
(4) CRITICAL RATIO (CR)

13. Example 8.2 – An example of priority rules
(1) Plane
t[i]
F[i]
makespan = 95
SPT = {2, 4, 3, 5, 1} = 49.6
(2) Plane
t[i]
# of
Passengers
# of Passengers
per minute
F[i]
# of
Passengers
{5, 1, 4, 3, 2}

14. (3) ARRIVAL TIME = DUE TIME.
100%
90%
% Passengers
76%
45% 23 49 65
time
(3) ARRIVAL TIME = DUE TIME.
(4) PRIORITY e.g.
continuing flights.
low fuel level.
carrying precious or perishable cargo.

15. ◇ § S9：EDD Scheduling § S10：Moore’s (1968) algorithm
minimizes the maximum lateness. ◇
§ S10：Moore’s (1968) algorithm
minimizes the number of Tardy jobs.
Step1：Sequence by earliest due date
i.e. d[1] ≦ d[2] ≦ d[3] ≦ …≦ d[n]
Step2：Find the 1st tardy job in the current
sequence, say job i. IF None exists go to Step 4
Step3：Consider jobs [1], [2], …, [i]
Reject the job with largest tj .and Return to Step2.
Step4：Current sequence + rejected job(In any order.)
Applications：chef, runway, preparing exam.
garage, dock.

16. Example 8.3 Ti = { max ( F[i] - D[i] , 0 ) } > 0 Job# 2 3 1 5 4 6
× 1 ×
Job# di ti Fi ×
× 5

17. Example 8.3 di ti Fi
Done!
1 - 5 di ti Fi

18. §. S10.1: Class Problems Discussion
Chapter 8 : # 3, 4, p , 32(a),(b), p
Preparation Time : 15 ~ 20 minutes Discussion : minutes

19. ◇ § S11：Lawler’s Algorithm : Precedence min. max. gi(Fi) 1 ≦ i ≦ N
gi(Fi) = Fi – di = Li min. max. Lateness.
gi(Fi) = max (Fi – di , 0) Tardiness. …
next to
LAST
LAST

20. ◇
Example 8.4
JOBS: 1 2 3
JOB ti di 4 5 6

21. ◇ Example 8.4 Total Processing time of all jobs is 15 JOBS 3 5 6
v = { 3, 5, 6 }
Total Processing time of all jobs is 15
min
iv
min
iv
{gi(Fi)}= {Fi-di}
= min {15-9 , , 15-7}
= min {6 , 4 , 8}= 4
∴ JOB # 5 is scheduled last. (6th)
JOBS 3
v = { 3, 6 } 6
Total Processing time is 13
min
iv
{gi(Fi)}= min {13-9 , 13-7}= min {4 , 6}= 4
∴ JOB # 3 is scheduled last. (5th)
So far, { … , 3 , 5}

22. ◇ Example 8.4 JOBS 2 v = { 2, 6 } 6 Total Processing time is 9 min iv
{gi(Fi)}= min {9-6 , 9-7}= 2
∴ JOB # 6 is scheduled last. (4th)
So far, { … , 6 , 3 , 5}
JOBS 2
v = { 2, 4 } 4
Total Processing time is 8
min
iv
{gi(Fi)}= min {8-6 , 8-7}= 1
∴ JOB # 4 is scheduled 3rd
∴ So far, { … , 4 , 6 , 3 , 5}
JOBS → ∴ { , 2 , 4 , 6 , 3 , 5}
JOBS → ∴ { 1 , 2 , 4 , 6 , 3 , 5} 2 1

23. ◇ ∴ maximum tardiness is 4 days. Example 8.4
Job# ti Fi di Ti (Tardiness)
∴ maximum tardiness is 4 days.

24. ◇ §. S12: Class Problems Discussion
Chapter 8 : # 6, 7, 8, p # 37, p.451
Preparation Time : 10 ~ 15 minutes Discussion : minutes

25. ◇ § S13：n job on m M/C’s … … … …
(1) n JOBS Must BE PROCESSED ON 1 M/C’s
n ……
1 ……..(n-2) (n-1) n
∴ there are n! possible ways.
(2) “n” JOBS Must BE PROCESSED ON “m” M/C’s
n …… n!
　　 　 .…… n!
…… n!
　　∴ there are (n!)m possible ways. #12 p.428
M/C
M/C１
M/C 2 … … … …
M/C m

26. § S14：Permutation schedules – characteristics
It provides better system performance in terms of both total flow time ( makespan ) and average flow time ( ).
mean IDLE time?
(B) SCHEDULING n Jobs on 2 M/C’s : if each Jobs must be processed in the order M/C-1 then M/C-2.
Results: The permutation schedule will minimize
“makespan” and minimizes “ ”.
(C) Theorem 8.2: (p.422)
The optimal solution for scheduling n jobs on 2 M/C’s is always a permutation schedule. F F

27. § S15：2 jobs on 2 M/C’s
All Possible Schedules for Two Jobs on Two Machines.
(B) Assumes that both jobs must be processed first on M/C#1 then on M/C#2.
makespan idle flow time 9 6 10
Machine I J
Machine I J
Machine J I
Machine J I
Machine J I
Machine I J
Machine I J
Machine J I

28. ◇ § S16：Permutation Schedules - Definition
＊Permutation Schedules ＝ Same Sequence on
both (all) M/C’s
Total # of permutation schedules is exactly n!

29. ◇ § S17： Johnson’s Rule 2 M/C’s TO MINIMIZE THE MAKESPAN
(1) Definition:
M/C-A
M/C-B
Jobs must be processed first on M/C-A
then M/C-B
Ai : Processing Time of Job i on M/C-A
Bi : Processing Time of Job i on M/C-B
Job i procedes Job i+1
if min (Ai , Bi+1) < min (Ai+1,Bi)
2 M/C’s

30. ◇ (2) Working Procedures for Johnson’s rule:
1. List the values of Ai & Bi in 2 columns.
2. Find the smallest remaining element in the 2 columns.
If it appears in column A then schedule that job next.
If it appears in column B then schedule that job last.
3. Cross off the jobs as they are scheduled.
Stop when all jobs have been scheduled.
An easy way to implement Johnson’s Rule

31. ◇
Example 8.5
Five jobs are to be scheduled on two machines. The processing times are
Jobs Machine A Machine B

32. Example 8.5 (A) (B) makespan=30
(A)
if using SPT in Ai then obtain the above
_____________________________________________________
IF BY JOHSON’S ROLE TO MINIMIZE THE MAKESPAN
OR TOTAL FLOW TIME.
Rule: Job i precedes job i+1 if MIN(Ai,Bi+1)<(Ai+1,Bi)
in order
A # # # # #1
B # # # # #1
(B)
makespan=30

33. § S18：n jobs on 3 M/C’s ◇
(A)Objective : To minimize total flow time (“make span.”)
it is still true that a permutation schedule
is optimal.
(B) But it is not necessarily optimal for the case of F’ .
(C) 3 M/C’s can be reduced to 2 M/C’s (then using Johnson’s Rule to solve it)
if min Ai≧max Bi
or min Ci≧max Bi
“either one” of these conditions be satisfied.
then define

34. ◇
Example 8.6
Consider the following job times for a three-machine problem. Assume that the jobs are processed in the sequence A-B-C
M/C’s
Jobs A B C
Min Ci=6
Max Bi=6
∴ Min Ci ≧ Max Bi
let

35. ◇ Example 8.6 Using Johnson’s Rule 5-1-4-2-3 M/C’s Jobs A’ B’ 1 9 13
Using Johnson’s Rule

36. ◇
Example 8.6

37. ◇ §. S18.1: Class Problems Discussion
Chapter 8 : # 13, p , p.451 ◆
Preparation Time : 10 ~ 15 minutes Discussion : minutes

38. ◇ § S19：more on 3 M/C’s problems and
If neither ﹛min Ai ≧ max Bi Nor min Ci ≧ max Bi ﹜
Then Using
and
Will usually give “REASONABLE” but
possibly “SUBOPTIMAL” result.
Simplifying 3 M/C’s problems M/C’s.
(B) TO MINIMIZE “MAKESPAN” OR “TOTAL FLOW TIME”
A PERMUTATION SCHEDULE IS OPTIMAL ON 3 M/C’s

39. ◇ § S20：Akers’ procedures for solving 2 Jobs on m M/C’s problems
Draw a cartesian coordinate system.
Job 1 on X-axis
Job 2 on Y-axis
mark off the operation times
on X & Y axis
(2) BLOCK OUT AREAS for each M/C’s
(3)Determine a path from origin to the end of the final block.
move:
The path with minimum vertical distance is “Optimal”.

40. Example 8.7 ◇
A regional manufacturing firm produces a variety of household products. One is a wooden desk lamp. Prior to packing, the lamps must be sanded, lacquered, and polished. Each operation requires a different machine. There are currently shipments of two models awaiting processing. The times required for the three operations for each of the two shipments are
JOB JOB 2
Oper Time Oper. Time
(A) Sanding A
(B) Lacquering B
(C) Polishing C

41. ◇ 2B 1A 2C 1B 1C 2A
Example 8.7
Fig p.426
maximize the diagonal movement = min. horizontal
vertical

42. ◇
Fig p.426
Example 8.7

43. ◇
Example 8.8

44. ◇ Example 8.8 C B D A A1 →B1 →C1 A2 A2 ↓ C1 D2 Job 2 (BOB)10 8 6 4 12 14 16 18
(16,14)
(16,11)
(7,11)
A1 →B1 →C1
A2 A2 ↓ C1 D2
C2←D1←D1 ←D2
C2 B2
Time=16+1+3=20 C
Job 2
(BOB) B D A
Job (Reggie)
A2 →D2 →B2 →C2 →C2 →C2 →C1 →D1
A1 A1 A1 B1
Time=16+4+3=23

45. ◇ §. S21: Class Problems Discussion Chapter 8 : # 15,16 p.428 39 p.452
Preparation Time : 25 ~ 30 minutes Discussion : minutes

46. ◇ The End § S22：Parallel processing on identical M/C’s.
Materials Management
2017/4/26 ◇
§ S22：Parallel processing on identical M/C’s.
SPT → minimize mean flow time.
LPT → minimize total flow time, or makespan.
( longest )
The End
Dr. Yuan-Shyi Peter Chiu

47. Materials Management
2017/4/26
Scheduling
Preview : Chap. 8 [ pp.413~442 ]
Problem: # 5, #32(a),(b),
#7, #8, #10,
#37, #40, #39,
#14, #15, #16
Dr. Yuan-Shyi Peter Chiu